\(A=\frac{99}{x^2-3x+13}=\frac{99}{x^2-3x+\frac{9}{4}+\frac{43}{4}}=\frac{99}{\left(x-\frac{3}{2}\right)^2+\frac{43}{4}}\)
Ta thấy: \(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{43}{4}\ge\frac{43}{4}\)
\(\Rightarrow\frac{1}{\left(x-\frac{3}{2}\right)^2+\frac{43}{4}}\le\frac{4}{43}\)\(\Rightarrow A\le\frac{396}{43}\)
Dấu "=" xảy ra khi \(\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x=\frac{3}{2}\)
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